Question

Let $N=p_1{p}_2{p}_3$ and $p_1,p_2,p_3$ are distinct prime number. If $\sum d=3Nor\sigma \left(N\right)=3N$. Show that ${\sum }_{i=1}^N\frac{1}{d_i}=3.$

Answer 1

$N=P_1{P}_2{P}_3$

Divisor of N are

$1,p_1,p_2,p_3,p_1{p}_2,p_2{p}_3,p_1{p}_3,p_1{p}_2{p}_3$

It is given $1+p_1+p_2+p_3+p_1{p}_2+p_2{p}_3+p_1{p}_3+p{p}_1{p}_2{p}_3=3N$

${\sum }_{i=1}^N\frac{1}{d_i}=1+\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}+\frac{1}{p_1}{p}_2+\frac{1}{p_1{p}_3}+\frac{1}{p_2{p}_3}+\frac{1}{p_1{p}_2{p}_3}$

$=\frac{p_1{p}_2{p}_3+p_2{p}_3+p_1{p}_3+p_1{p}_2+p_3+p_2+p_1+1}{p_1{p}_2{p}_3}$

${\sum }_{i=1}^N\frac{1}{d_i}=\frac{3N}{N}=3$